Transcript Electromagnetic waves - Review Sandra Cruz-Pol, Ph. D.

Electromagnetic waves
-ReviewSandra Cruz-Pol, Ph. D.
ECE UPRM
Mayagüez, PR
Electromagnetic Spectrum
Cruz-Pol, Electromagnetics
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Maxwell Equations
in General Form
Differential form Integral Form
  D  v
 D  dS   v dv
s
 B  0
v
 B  dS  0
s
B
 E  
t

L E  dl   t s B  dS
D
 H  J 
t
D 

H

dl

J


L
s  t   dS
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Gauss’s Law for E
field.
Gauss’s Law for H
field. Nonexistence
of monopole
Faraday’s Law
Ampere’s Circuit
Law
Would magnetism would produce
electricity?

Eleven years later,
and at the same time,
Mike Faraday in
London and Joe
Henry in New York
discovered that a
time-varying magnetic
field would produce
an electric current!
Vemf
d
 N
dt

L E  dl   t s B  dS
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Electromagnetics was born!

This is the principle of
motors, hydro-electric
generators and
transformers operation.
This is what Oersted discovered
accidentally:
D 

L H  dl  s  J  t   dS
*Mention some examples of em waves
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Special case

Consider the case of a lossless medium

with no charges, i.e. .  v  0
 0
The wave equation can be derived from Maxwell
equations as
 E    c E  0
2
2
What is the solution for this differential equation?
 The equation of a wave!
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Phasors for harmonic fields
Working with harmonic fields is easier, but
requires knowledge of phasor.
 The phasor is multiplied by the time factor, ejt,
and taken the real part.
  t  
j
Re{re }  r cos(t   )
j
Im{re }  r sin( t   )
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Advantages of phasors
 Time
derivative is equivalent to
multiplying its phasor by j
A
 jAs
t
 Time
integral is equivalent to dividing by
the same term.
As
 At  j
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Time-Harmonic fields
(sines and cosines)

The wave equation can be derived from Maxwell
equations, indicating that the changes in the
fields behave as a wave, called an
electromagnetic field.

Since any periodic wave can be represented as
a sum of sines and cosines (using Fourier), then
we can deal only with harmonic fields to simplify
the equations.
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Maxwell Equations
for Harmonic fields
Differential form*

 DE
 v  v
Gauss’s Law for E field.
 BH
0

Gauss’s Law for H field.
No monopole
0
  E   jH
 E  
B
t
  H  J  jE
D
 H  J 
t
* (substituting
Faraday’s Law
Ampere’s Circuit Law
D   E andCruz-Pol,
H  Electromagnetics
B)
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A wave

Start taking the curl of Faraday’s law
    Es   j  H s

Then apply the vectorial identity
    A  (  A)   2 A

And you’re left with
(  Es )   Es   j (  j ) Es
2
  Es
2
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A Wave
 E  E  0
2
2
Let’s look at a special case for simplicity
without loosing generality:
•The electric field has only an x-component
•The field travels in z direction
Then we have
E ( z, t )
whose general solution is
E(z)  Eo e z  Eo' e z
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To change back to time domain

From phasor
E xs ( z )  Eo e

 z
 Eo e
 z (  j )
…to time domain
E ( z , t )  Eo e
 z

cos(t  z ) x
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Several Cases of Media
1.
2.
3.
4.
Free space (  0,    o ,    o )
Lossless dielectric (  0,    r  o ,    r  o or    )
Lossy dielectric (  0,    r  o ,    r  o )
Good Conductor (  ,    o ,    r o or    )
Permitivity:
o=8.854 x 10-12[ F/m]
Permeability: o= 4p x 10-7 [H/m]
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1. Free space
There are no losses, e.g.

E ( z , t )  A sin( t  z ) x
Let’s define
 The phase of the wave
 The angular frequency
 Phase constant
 The phase velocity of the wave
 The period and wavelength
 How does it moves?
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3. Lossy Dielectrics
(General Case)

In general, we had

E ( z , t )  Eo e z cos(t  z ) x
 2  j (  j )
    j
 Re  2   2   2   2 
 2   2   2    2   2 2

From this we obtain
2
2





 

 


 1 
 1 
 
  1 and   
  1
2 
 
2 
 









So , for a known material and frequency, we can find j
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Intrinsic Impedance, h

If we divide E by H, we get units of ohms and
the definition of the intrinsic impedance of a
medium at a given frequency.
j
 h h
h
  j
E ( z, t )  Eo e
H ( z, t ) 
Eo
h
z

cos(t  z ) x
e z cos(t  z  h ) yˆ
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[]
*Not in-phase
for a lossy
medium
Note…
E ( z, t )  Eo e
H ( z, t ) 
Eo
h
z

cos(t  z ) x
e z cos(t  z  h ) yˆ

E and H are perpendicular to one another
 Travel is perpendicular to the direction of
propagation
 The amplitude is related to the impedance
 And so is the phase
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Loss Tangent

If we divide the conduction current by the
displacement current
J cs
J ds
 Es



 tan   loss tangent
j Es 
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Relation between tan and c
 

  H  E  j E  j 1  j
E

 

 j c E
The complex permittivi ty is
 

 c   1  j    ' j ' '
 

" 
The loss tangent can be defined also as tan   
 ' 
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2. Lossless dielectric
(  0,    r  o ,   r o or    )
 Substituting
in the general equations:
  0,    

1
2p
u 




 o
h
0

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Review: 1. Free Space
(  0,    o ,   o )

Substituting in the general equations:
  0,       / c

1
2p
u 
c



 o o
h
o o
0  120p   377 
o

E ( z , t )  Eo cos(t  z ) x V / m
H ( z, t ) 
Eo
ho
cos(t  z ) yˆ
A/ m
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4. Good Conductors
(  ,    o ,   r o )

Substituting in the general equations:
  

2

2
u 



2p
Is water a good
conductor???


h
45o


E ( z , t )  Eo e z cos(t   z ) x [V / m]
H ( z, t ) 
Eo
e z cos(t  z  45o ) yˆ [ A / m]

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Skin depth, d

Is defined as the
depth at which the
electric amplitude is
decreased to 37%
1
e  0.37  (37%)
e
z
e
1
at z  1 /   d
d  1 /  [m]
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Short Cut …

You can use Maxwell’s or use
 1

H  kˆ  E
h


E  h kˆ  H
where k is the direction of propagation of the wave,
i.e., the direction in which the EM wave is
traveling (a unitary vector).
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Exercises: Wave Propagation in
Lossless materials

A wave in a nonmagnetic material is given by

H  zˆ50 cos(109 t  5 y ) [mA/m]
Find:
(a) direction of wave propagation,
(b) wavelength in the material
(c) phase velocity
(d) Relative permittivity of material
(e) Electric field phasor

 j5 y
[V/m]
Answer: +y, up= 2x108 m/s, 1.26m, 2.25, E   xˆ12.57e
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Power in a wave

A wave carries power and transmits it
wherever it goes
The power density per
area carried by a wave
is given by the
Poynting vector.
See Applet by Daniel RothCruz-Pol,
at
Electromagnetics
UPRM
http://www.netzmedien.de/software/download/java/oszillator/
Poynting Vector Derivation…
  2  2
2


E

H

dS


E

H
dv


E
dv


S


t v  2
2

v
Total power across
surface of volume

Rate of change of
stored energy in E or H
Ohmic losses due to
conduction current
Which means that the total power coming out of
a volume is either due to the electric or
magnetic field energy variations or is lost as
ohmic losses.
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Power: Poynting Vector

Waves carry energy and information
 Poynting says that the net power flowing out of a
given volume is = to the decrease in time in
energy stored minus the conduction losses.
   
2
S  P  E  H [W/m ]
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Represents the
instantaneous
power vector
associated to the
electromagnetic
wave.
Time Average Power

The Poynting vector averaged in time is



T
T 

 *


1
1
1
Pave   P dt   E  H dt  Re Es  H s
T0
T0
2


For the general case wave:
Es  Eo e z e  jz xˆ [V / m]
Hs 
Eo
h
e z e  jz yˆ [ A / m]
Pave
Eo2  2z

e
cos h zˆ
2h
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[W/m 2 ]
Total Power in W
The total power through a surface S is

Pave   Pave  dS [W ]
S



Note that the units now are in Watts
Note that power nomenclature, P is not cursive.
Note that the dot product indicates that the surface
area needs to be perpendicular to the Poynting
vector so that all the power will go thru. (give example
of receiver antenna)
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Exercises: Power
1. At microwave frequencies, the power density considered
safe for human exposure is 1 mW/cm2. A radar radiates
a wave with an electric field amplitude E that decays
with distance as E(R)=3000/R [V/m], where R is the
distance in meters. What is the radius of the unsafe
region?
 Answer: 34.6 m
2. A 5GHz wave traveling in a nonmagnetic medium with
r=9 is characterized by E  yˆ 3 cos(t  x)  zˆ 2 cos(t  x)[V/m]
Determine the direction of wave travel and the average
power density carried by the wave

2
 Answer: Pave   xˆ 0.05 [W/m ]
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TEM wave
x
z
x
z
y
Transverse ElectroMagnetic = plane wave
 There are no fields parallel to the direction of
propagation,
 only perpendicular (transverse).
 If have an electric field Ex(z)


…then must have a corresponding magnetic field
Hx(z)
The direction of propagation is
aˆ E  aˆ H  aˆk
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Polarization of a wave
IEEE Definition:
The trace of the tip of the
E-field vector as a
function of time seen from
behind.
x
x
Simple cases
 Vertical, Ex
Exs ( z )  Eo e
y
y
x
 jz
Ex ( z )  Eo cos(t  z ) xˆ

Horizontal, Ey
y
y
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x
Polarization:
Why do we care??
 Antenna applications –


Remote Sensing and Radar Applications –


Antenna can only TX or RX a polarization it is designed to support.
Straight wires, square waveguides, and similar rectangular systems
support linear waves (polarized in one direction, often) Circular
waveguides, helical or flat spiral antennas produce circular or
elliptical waves.
Many targets will reflect or absorb EM waves differently for different
polarizations. Using multiple polarizations can give different
information and improve results. Rain attenuation effect.
Absorption applications –

Human body, for instance, will absorb waves with E oriented from
head to toe better than side-to-side, esp. in grounded cases. Also,
the frequency at which maximum absorption occurs is different for
these two polarizations. This has ramifications in safety guidelines
and studies.
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Polarization

In general, plane wave has 2 components; in x & y
E ( z )  xˆE x  yˆE y

And y-component might be out of phase wrt to xcomponent, d is the phase difference between x and y.
E x  Eo e
 j z
x
E y  E o e  j z  d
Ex
x
y
y
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Ey
Front View
Several Cases
polarization: ddy-dx =0o or ±180on
 Circular polarization: dy-dx =±90o & Eox=Eoy
 Elliptical polarization: dy-dx=±90o & Eox≠Eoy,
or d≠0o or ≠180on even if Eox=Eoy
 Unpolarized- natural radiation
 Linear
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Linear polarization

Front View
d =0
x
E x  E o e  j z
Ex
E y  E o e  j z

y
Ey
@z=0 in time domain
E x  E xo cos(t)
E y  E yo cos(t)
Back View:
x
y
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Circular polarization

Both components have
same amplitude Eox=Eoy,
 d =d y-d x= -90o = Right
circular polarized (RCP)
 d =+90o = LCP
E x  E xo cos(t)
E y  E yo cos(t  90o )
in phasor :
E x  xˆE xo  yˆ E yo e  j 90  xˆE xo  jEyo yˆ
x
y
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Elliptical polarization

X and Y components have different amplitudes
Eox≠Eoy, and d =±90o

Or d ≠±90o and Eox=Eoy,
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Polarization example
All light
comes out
Unpolarized
radiation
enters
Nothing comes
out this time.
Polarizing glasses
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sin(   90o )  cos( )
sin(   180 )  sin(  )
cos(  90 )  sin ( )
cos(  180 )   cos( )
Example
o

o
o
Determine the polarization state of a plane wave
with electric field:
a. E ( z, t )  xˆ3cos(t - z  30o ) - ŷ4sin( t - z  45o )
o
o
b.E ( z, t )  xˆ3cos(t - z  45 )  ŷ8sin( t - z  45 )
o
o
ˆ
E
(
z
,
t
)

x
4
cos(

t

z

45
)
ŷ
4sin(

t

z

45
)
c.
-jy
ˆ
ˆ
d. Es ( y )  14( x-jz )e
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a.
Elliptic
b.
-90, RHEP
c.
LP<135
d.
-90, RHCP
Cell phone & brain

Computer model for
Cell phone Radiation
inside the Human
Brain
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Decibel Scale

In many applications need comparison of two
powers, a power ratio, e.g. reflected power,
attenuated power, gain,…
 The decibel (dB) scale is logarithmic
G
Pout
Pin
 Vo 2 /R 
 Pout 
V 
  10 log  2   20 log  o 
G[dB]  10 log 
 V /R 
 Pin 
 Vi 
 i


Note that for voltages, the log is multiplied by 20
instead of 10.
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Attenuation rate, A

Represents the rate of decrease of the magnitude
of Pave(z) as a function of propagation distance
 Pave(z) 
  10 log e 2z
A  10 log 
 Pave( 0 ) 
 20z log e  - 8.68z  - dB z [dB]

where
 dB[dB/m ]  8.68 [ Np/m]
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
Summary
Any medium


uc

Low-loss medium
(”/’<.01)
0
 
2

 

 1 
  1
2 







h
Lossless
medium
(=0)



2
Good conductor
(”/’>100)
Units
[Np/m]
pf
[rad/m]
 
 
pf

(1  j )

j
  j


/
1
1
4pf



2p/up/f
**In free space;


up
up
up
f
f
f
-12 F/m
Cruz-Pol,10
Electromagnetics
o =8.85
o=4p 10-7 H/m
UPRM
[ohm]
[m/s]
[m]
Exercise: Lossy media
propagation
For each of the following determine if the material is low-loss dielectric,
good conductor, etc.
(a) Glass with r=1, r=5 and =10-12 S/m at 10 GHZ
(b) Animal tissue with r=1, r=12 and =0.3 S/m at 100 MHZ
(c) Wood with r=1, r=3 and =10-4 S/m at 1 kHZ
Answers:
(a)
(b)
(c)
low-loss,  8.4x1011 Np/m,  468 r/m,  1.34 cm, up1.34x108, hc168 
general,  9.75, 12, 52 cm, up0.5x108 m/s, hc39.5j31.7 
Good conductor,  6.3x104,  6.3x104,  10km, up0.1x108, hc6.281j 
Cruz-Pol, Electromagnetics
UPRM
Reflection and Transmission
Wave incidence

Wave arrives at an angle



Snell’s Law and Critical angle
Parallel or Perpendicular
Brewster angle
Cruz-Pol, Electromagnetics
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EM Waves
Medium 2 : 2, 2
Medium 1 : 1 , 1
kiz
kix
ki
i
r



Normal , an
Plane of
incidence
Angle of
incidence
kr
y
z=0
Cruz-Pol, Electromagnetics
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t
kt
Property
Reflection
coefficient
Transmission
coefficient
Relation
Power
Reflectivity
Power
Transmissivity
Snell’s Law:
Normal
Incidence
Perpendicular
h 2  h1

h 2  h1
 
2h 2 i

h 2  h1t
 
h 2 cos  i  h1 cos  t
h 2 cos  i  h1 cos  t
  1  
R 
R  
T  1 R
n
sin  t  sin 
n
|| 
h 2 cos  t  h1 cos  i
h 2 cos  t  h1 cos  i
2h 2 cos  i
2h 2 cos  i


h 2 cos  i  h1 cos  t || h cos   h cos 
2
t
1
i
 1  
2
Parallel
T  1  R
1
Cruz-Pol,
Electromagnetics
i
UPRM
2
where
 ||  1  || 
2
cos  i
cos  t
R||  ||
2
T||  1  R||
n2   r 2 r 2
Critical angle, c
…All is reflected
When t =90o, the refracted
wave flows along the surface
and no energy is transmitted
into medium 2.
 The value of the angle of
incidence corresponding to
this is called critical angle, c.
 If i > c, the incident wave is
totally reflected.

Cruz-Pol, Electromagnetics
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n2
sin  c  sin  t [ t  90 o ]
n1
n2
r2


n1
 r1
(for 1   2 )
Fiber optics
 Light
can be guided with total reflections
through thin dielectric rods made of
glass or transparent plastic, known as
optical fibers.
 The
only power lost is due to reflections
at the input and output ends and
absorption by the fiber material (not
perfect dielectric).
Cruz-Pol, Electromagnetics
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Optical fibers have cylindrical fiber core with index of
refraction nf, surrounded by another cylinder of
lower, nc < nf , called a cladding.
[Figure from Ulaby, 1999]

For total reflection:
n2
sin  3  sin  c 
 2   3  90o
n1
Cruz-Pol, Electromagnetics
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Acceptance angle
sin  a 
(n 2f  nc2 )
n0
Brewster angle, B

Is defined as the incidence angle at which the
reflection coefficient is 0 (total transmission).
h 2 cos  t  h1 cos  B
|| 
0
h 2 cos  t  h1 cos  B
h 2 cos  t  h1 cos  B  0
1  ( 1 2 /  2 1 )
sin  B|| 
1  ( 1 /  2 ) 2

* B is
known as
the
polarizing
angle
The Brewster angle does not exist for perpendicular
polarization for nonmagnetic materials.
Cruz-Pol, Electromagnetics
http://www.amanogawa.com/archive/Oblique/Oblique-2.html
UPRM
Reflection vs. Incidence angle.
Reflection vs.
incidence angle
for different
types of soil and
parallel or
perpendicular
polarization.
Cruz-Pol, Electromagnetics
UPRM
Antennas
 Now
let’s review antenna theory
Cruz-Pol, Electromagnetics
UPRM